Question

the dimensions of a room are 15M,10M and 8M the volume of a bag is 2.25 M^3 the maximum number of bags that can be accommodate in the room​

Answer 1

Answer:

Around 533.33 bags

Step-by-step explanation:

total no. of bags = n

Vol. of one bag = 2.25m³

Vol. of room = Length×Breadth×Height

= 15m×10m×8m

= 1200m³

total no. of bags to be accommodated= n×volume of bag = volume of room

n×2.25m³= 1200m³

n=1200m³÷2.25m³

m³ will be cancelled

n= 533.33 bags

Answer 2

Answer:

Around 533.33 bags

Step-by-step explanation:

Given:

To find:

total no. of bags = n

Vol. of one bag = 2.25m³

$Vol. of room = Length*Breadth*Height$

$= 15m*10m*8m$

$= 1200\ m^3$

total no. of bags to be accommodated= n× volume of bag = volume of room

n×2.25m³= 1200m³

n=1200m³÷2.25m³

m³ will be cancelled

n= 533.33 bags